Approximate the probability that $100$ elements, each of which works for time $T_{i}$, will provide $100$ hours of work in total. It is known that $E[T_{i}]=1$ and $D^2[T_{i}]=1$.
I have introduced a new variable $$Z=\sum_{i=1}^{100}T_i$$ so that $$E[Z]=E[nT_i]=n$$ $$D^2[Z]=D^2[nT_i]=n^2D^2[T_i]=n^2$$ where $n=100$.
I'm not sure about the following steps: $$Pr(\frac{nT_i-n\cdot n}{n \sqrt{n}}\geqslant\frac{100-n\cdot n}{n\sqrt{n}})=1-Pr(\frac{nT_i-n\cdot n}{n\sqrt{n}}<\frac{100-n\cdot n}{n\sqrt{n}})=1-\Phi(\frac{100-n\cdot n}{n\sqrt{n}})$$
Is my solution correct? I'm really not sure about this one.
There is no question in your exercise.
I do not know what do you mean with $D^2$. If that is the variance, usually indicated with $V(X)$ or $\sigma^2_{X}$,
and assuming that the question is to calculate the probability $\mathbb{P}[\sum_i X_i>100]$ you can apply the CLT that states:
$$\frac{\sum_i X_i-n\mu}{\sigma \sqrt{n}}$$
Your probability is
$$\mathbb{P}[Z>\frac{100-100\cdot 1}{1\cdot \sqrt{100}}]=1-\Phi(0)=\frac{1}{2}$$
Note:
As per the fact that $\mathbb{E}[\sum_i X_i]=n\mu$ the result is $\frac{1}{2}$ without applying the CLT